Question

Tìm x biết rằng

\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+...+\(\frac{1}{x\left(x+3\right)}\)=\(\frac{101}{1540}\)

Answer 1

\(\frac{1}{5.8}+\frac{1}{8.11}+...........+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+............+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+............+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

Answer 2

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(\frac{1}{x\left(x+3\right)}=\frac{1}{x}-\frac{1}{x+3}\)

VT: 2\(\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}....\frac{1}{x+3}\right)\)

Answer 3

cảm ơn các bn nhìu

Answer 4

\(M=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+3}\right)\)

\(M=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)\)

Thay M vào ta có :

\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

=> \(\left(\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}:\frac{1}{3}\)

=> \(\frac{1}{1}-\frac{1}{x+3}=\frac{303}{1540}\)

=> \(\frac{1}{x+3}=\frac{1}{1}-\frac{303}{1540}\)

=> \(\frac{1}{x+3}=\frac{1237}{1540}\)

=> x+3 = 1540

=> x = 1537

Answer 5

Đặt A=vế trái

=>3A=3*101/1540

<=> 3/5.8 + 3/8.11 + 3/11.14 + ...+ 3/x.(x+3)=303/1540

<=> 1/5 -1/8+1/8-1/11+1/11-1/14+...+1/x-1/(x+3)=303/1540

<=> 1/5-1/(x+3)=303/1540 => LÀ được

Answer 6

ý ả nhầm đề

Answer 7

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(\frac{1}{3}\cdot\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\cdot\left(x+3\right)}\right)=\frac{101}{1540}\)

\(\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{308}\)

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

( Mình ko copy cùa ai hết )

Answer 8

1/3.(3/5.8+3/8.11+3/11.14+...+3/x.(x+3))=101/1540

1/3.(1/5-1/8+1/8-1/11+1/11-1/14+.....+1/x-1/x+3)=101/1540

1/3.(1/5-1/x+3)=101/1540

1/5-1/x+3=101/1540:1/3

1/5-1/x+3=303/1540

1/x+3=1/5-303/1540

1/x+3=1/308

Suy ra x+3=308

x=308-3=305